在XAML中设置WPF ContextMenu的PlacementTarget属性?

您应该在按钮上设置contextmenuservice.placement附加属性,如文档中的备注所述 ContextMenu.Placement .

<Button Name="btnFoo" Content="Foo" ContextMenuService.Placement="Bottom">
    <Button.ContextMenu>
        <ContextMenu>
            <MenuItem Header="Bar" />
        </ContextMenu>
    </Button.ContextMenu>
</Button>

您是否尝试过:

<Button Name="btnFoo" Content="Foo">
    <Button.ContextMenu>
        <ContextMenu>
            <MenuItem Header="Bar" />
        </ContextMenu>
    </Button.ContextMenu>
</Button>

这将使ContextMenu在右键单击鼠标的位置(在按钮上)打开。 我认为这可能是你想要的位置,对吗?

---编辑--- 在这种情况下,请使用:

<Button Name="btnFoo" Content="Foo" ContextMenuOpening="ContextMenu_ContextMenuOpening">
    <Button.ContextMenu>
        <ContextMenu Placement="Bottom">
            <MenuItem Header="Bar" />
        </ContextMenu>
    </Button.ContextMenu>
</Button>

在代码背后:

private void ContextMenu_ContextMenuOpening(object sender, ContextMenuEventArgs e)
{
    // Get the button and check for nulls
    Button button = sender as Button;
    if (button == null || button.ContextMenu == null)
        return;
    // Set the placement target of the ContextMenu to the button
    button.ContextMenu.PlacementTarget = button;
    // Open the ContextMenu
    button.ContextMenu.IsOpen = true;
    e.Handled = true;
}

您可以为多个按钮和ContextMenu重用该方法。

你可以用一个 <Menu /> ,样式为 Button 并避免与 ContextMenuService .